Question

Write the electron configuration for the first excited state for \(\mathrm{N}_{2}\), that is, the state with the highest-energy electron moved to the next available energy level. (a) Is the nitrogen in its first excited state diamagnetic or paramagnetic? (b) Is the \(\mathrm{N}-\mathrm{N}\) bond strength in the first excited state stronger or weaker compared to that in the ground state?

Short answer

The electron configuration for the first excited state of N2 is σ(1s)² < σ*(1s)² < σ(2s)² < σ*(2s)² < π(2p)⁴ < σ(2p)¹ < π*(2p)¹ < σ*(2p)0. (a) The nitrogen in its first excited state is paramagnetic due to the presence of unpaired electrons in the σ(2p) and π*(2p) orbitals. (b) The N-N bond strength in the first excited state is weaker compared to that in the ground state, as the bond order in the first excited state (2) is lower than that in the ground state (3).

Step-by-step solution

Determine the electron configuration for the ground state of N2

To determine the electron configuration, we should know that Nitrogen has 7 electrons, so N2 will have a total of 14 electrons. Nitrogen belongs to the second period in the periodic table, so its electron configuration can be written as: \(1s^2 2s^2 2p^3\). Since we have two nitrogen atoms, we need to fill the molecular orbitals (MOs) according to the Aufbau principle (from lower to higher energy levels) and Hund's rule (maximize the unpaired electrons). The MO energy levels for N2 are: σ(1s) < σ*(1s) < σ(2s) < σ*(2s) < π(2p) < σ(2p) < π*(2p) < σ*(2p) Now let's fill the MOs with 14 electrons: σ(1s)² < σ*(1s)² < σ(2s)² < σ*(2s)² < π(2p)⁴ < σ(2p)² < π*(2p)0 < σ*(2p)0

Determine the electron configuration for the first excited state of N2

For the first excited state, we will move the highest-energy electron to the next available energy level: σ(1s)² < σ*(1s)² < σ(2s)² < σ*(2s)² < π(2p)⁴ < σ(2p)¹ < π*(2p)¹ < σ*(2p)0

Determine if the nitrogen in its first excited state is diamagnetic or paramagnetic

Diamagnetic substances have all of their electrons paired, while paramagnetic substances have at least one unpaired electron. In our case, we have unpaired electrons in the σ(2p) and π*(2p) orbitals: σ(1s)² < σ*(1s)² < σ(2s)² < σ*(2s)² < π(2p)⁴ < σ(2p)¹ < π*(2p)¹ < σ*(2p)0 So, the nitrogen in its first excited state is paramagnetic.

Compare the N-N bond strength in the first excited state with that in the ground state

The bond order is given by the formula: Bond order = (Number of electrons in bonding orbitals - Number of electrons in antibonding orbitals) / 2 For the ground state: Bond order = (10 - 4) / 2 = 3 For the first excited state: Bond order = (9 - 5) / 2 = 2 The bond strength is directly proportional to the bond order. Since the bond order in the first excited state (2) is lower than that in the ground state (3), the N-N bond strength in the first excited state is weaker compared to that in the ground state.

Key concepts

Molecular Orbitals

Understanding molecular orbitals (MOs) is vital for analyzing chemical bonding in molecules. MOs are formed when atomic orbitals, which are the regions around an individual atom where electrons are likely to be found, overlap between two or more atoms in a molecule.

According to the MO theory, these orbitals can be categorized as bonding, antibonding, or non-bonding. Bonding orbitals lower the energy of the system and hold the atoms together, as they are filled with pairs of electrons from each atom. In contrast, electrons in antibonding orbitals can weaken relationships due to their higher energy.

The filling of these orbitals in diatomic nitrogen (2) follows the principles of Aufbau (filling from lowest energy level), Pauli Exclusion (each orbital hosts at most two electrons with opposite spins), and Hund's (maximize unpaired electrons before pairing up). When we move an electron to create an 'excited state', we shift an electron from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). This process, as shown in the exercise, changes the electron configuration and often the magnetic properties and bonding strength within the molecule.

Paramagnetic vs Diamagnetic

The concepts of paramagnetism and diamagnetism describe how substances react in magnetic fields, which is largely based on the electron configurations within their atoms or molecules.

Materials that are diamagnetic have all their electrons paired up within orbitals, leading to a cancelation of their magnetic moments. Thus, diamagnetic substances do not exhibit attraction to magnetic fields and might even be repelled by them. On the other hand, paramagnetic substances have unpaired electrons, which imbue them with temporary magnetic moments when they are in the presence of a magnetic field, resulting in attraction towards it.

The nitrogen molecule in its first excited state, as indicated by the solution, shows paramagnetism due to the presence of unpaired electrons in the π*(2p) and σ(2p) orbitals. This fundamental concept helps to understand and predict the behavior of substances in various chemical and physical contexts.

Bond Order Calculation

Bond order is a mathematical representation used to give an indication of the strength and stability of a bond in molecular orbital theory. It is determined by taking the difference in the number of electrons occupying bonding and antibonding orbitals divided by two.

Mathematically, it can be expressed as:

[Bond order = \(\frac{(Number \ of \ electrons \ in \ bonding \ orbitals - Number \ of \ electrons \ in \ antibonding \ orbitals)}{2}\)]A higher bond order typically suggests a stronger and more stable bond. As the exercise reveals, the bond order for the ground state of 2 is 3, indicating a very strong triple bond. In the first excited state, the bond order drops to 2 due to the shifting of an electron to an antibonding orbital, reflecting a decrease in bond strength to that of a double bond. This demonstrates how excitation can affect molecular stability, which is critical for students to grasp when studying chemical reactions and molecular dynamics.